Chapter 6, Problem 6.170AP

Interpretation Introduction

Interpretation: The rate of effusion of an unknown gas is to be given $0.10\text{m/s}$ and under identical experimental conditions the rate of effusion of ${\text{SO}}_{\text{3}}(\text{g})\text{is}0.052\text{m/s}$ given. The molar mass of unknown gas is to be calculated.

Concept introduction: The process in which the gas is escapes through a tiny hole from its container into the region where pressure is low is called as effusion. Graham’s gives a relationship that the rate of effusion of gases is inversely proportional to the square root of their molar masses. This relationship is known as Graham’s law of effusion.

To determine: The molar mass of unknown gas.